Use the fact that the variance of a Poisson distribution is σ2=μ. The mean number of bankruptcies filed per hour by businesses in a country was about five. ​(a) Find the variance and the standard deviation. Interpret the results.​ (b) Find the probability that at most three businesses will file bankruptcy in any given hour.

Answers

Answer 1

Answer:

Variance = 5

Standard deviation = 2.236

0.2650257

Step-by-step explanation:

For a Poisson distribution is σ² = μ.

Given that:

Mean , μ of bankruptcies files per hour = 5

μ = 5

For a Poisson distribution :

P(x = x) = (μ^x * e^-μ) / x!

The Variance and standard deviation :

Variance : σ² = μ = 5

Standard deviation = sqrt(variance) = sqrt(5) = 2.236

B.) Find the probability that at most three businesses will file bankruptcy in any given hour.

P( x ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(x = 0) = (5^0 * e^-5) / 0! = 0.0067379

P(x = 1) = (5^1 * e^-5) / 1! = 0.0336897

P(x = 2) = (5^2 * e^-5) / 2! = 0.0842243

P(x = 3) = (5^3 * e^-5) / 3! = 0.1403738

0.0067379 + 0.0336897 + 0.0842243 + 0.1403738

= 0.2650257

Answer 2

Using the Poisson distribution, it is found that:

a) The variance is of 5 while the standard deviation is of 2.24. It means that over many hours, the number of bankruptcies should be within 2.24 of 5.

b) 0.265 = 26.5% probability that at most three businesses will file bankruptcy in any given hour.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

The parameters are:

x is the number of successes e = 2.71828 is the Euler number [tex]\mu[/tex] is the mean in the given interval, which is the same as the variance.

In this problem, the mean is of [tex]\mu = 5[/tex].

Item a:

The variance is the same as the mean, of 5.The standard deviation is the square root of the variance, hence [tex]\sqrt{5} = 2.24[/tex]

The variance is of 5 while the standard deviation is of 2.24. It means that over many hours, the number of bankruptcies should be within 2.24 of 5.

Item b:

This probability is:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

In which

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-5}(5)^{0}}{(0)!} = 0.0067[/tex]

[tex]P(X = 1) = \frac{e^{-5}(5)^{1}}{(1)!} = 0.0337[/tex]

[tex]P(X = 2) = \frac{e^{-5}(5)^{2}}{(2)!} = 0.0842[/tex]

[tex]P(X = 3) = \frac{e^{-5}(5)^{3}}{(3)!} = 0.1404[/tex]

Then:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0067 + 0.0337 + 0.0842 + 0.1404 = 0.265[/tex]

0.265 = 26.5% probability that at most three businesses will file bankruptcy in any given hour.

A similar problem is given at https://brainly.com/question/16912674


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